Electric fields and capacitors

[Ras9]

Hi guys, I was tutoring some students and I was explaining them how to calculate the capacitance between two parallels planes. They made me some questions and actually I am not quite sure about my answers!

I started by showing them how to calculate the electric field generated by a single infinite plane whit homogeneous charge distribution. The field is E_0={\sigma \over 2 \epsilon_0}. The first question regarded the dependence of this field with this distance. Is it constant at all distances? Also at infinite? I assumed so, since the plane itself is infinite.

Then we moved to the case with two planes with opposite charges distribution. The field in this case would just be E_0={\sigma \over \epsilon_0} and so the capacitance C={Q \over \Delta V} =\epsilon_0{S \over d} where S and d are the surfaces of the plane and the distance between the two. To obtain this result, though, I used the electric field generated by an infinite plane, and then I assumed che charge distribution to be the ratio over the total charge and the surface. What did I miss in the passage from an infinite plane to a finite one? Is the electric field generated by a finite plane equal to the one generated by an infinite one? If so how does one answer to the first question?

The last part regarded spherical capacitors. To calculate its capacitance one in general uses only the electric field due to the charge present in the inner capacitor? Why is there no contribution from the opposite charge present in the external one?

Thanks for the help!

 

[tech99]

Hi guys, I was tutoring some students and I was explaining them how to calculate the capacitance between two parallels planes. They made me some questions and actually I am not quite sure about my answers!

I started by showing them how to calculate the electric field generated by a single infinite plane whit homogeneous charge distribution. The field is E_0={\sigma \over 2 \epsilon_0}. The first question regarded the dependence of this field with this distance. Is it constant at all distances? Also at infinite? I assumed so, since the plane itself is infinite.

Then we moved to the case with two planes with opposite charges distribution. The field in this case would just be E_0={\sigma \over \epsilon_0} and so the capacitance C={Q \over \Delta V} =\epsilon_0{S \over d} where S and d are the surfaces of the plane and the distance between the two. To obtain this result, though, I used the electric field generated by an infinite plane, and then I assumed che charge distribution to be the ratio over the total charge and the surface. What did I miss in the passage from an infinite plane to a finite one? Is the electric field generated by a finite plane equal to the one generated by an infinite one? If so how does one answer to the first question?

The last part regarded spherical capacitors. To calculate its capacitance one in general uses only the electric field due to the charge present in the inner capacitor? Why is there no contribution from the opposite charge present in the external one?

Thanks for the help!

I think you have used Charge per Unit Area in your formula, so the plate can change in size as wanted. We assume that the field lines are perpendicular to the surface, so they are parallel and not diverging. For a finite plate, this means that the distance must be not too large, or else the lines will spread out, as from a point source. For a plate, we have field lines on both front and back by the way. For the spherical capacitor, we assume that the charges all reside on the outside surface of the spheres (Faraday's Ice Pail Experiment etc). So there is no charge on the inner surface of the outer sphere.

 

[Delta2]

It all depends on some assumption you make on how the field will look like (due to some sort of symmetry that is present) and with the help of that assumption to apply Gauss's Law and to solve for the electric field.

So in the case of an infinite plate it is easy to see the assumption that all the points at a distance x from the infinite plate will have the same electric field which will be also perpendicular to the infinite plate (because the plate is infinite all the points at distance x are symmetric to the plate). The points at distance -x (i.e on the other side of infinite plate) will have same electric field in magnitude and opposite direction. From this assumption and with Gauss's Law you can conclude that the electric field will , at the end, be independent of distance x and equal to $\sigma /2\epsilon_0$

In the case of two finite plates at distance d (i.e a parallel plate capacitor) we assume that the field will be confined only in the space between the two plates and it will have direction perpendicular to the surface of the plates. This assumption is true if the distance d between the plates is relatively small compared to the other 2 finite dimensions of the finite plates. With this assumption we can conclude that at a distance x from the left plate all points will have the same electric field in magnitude and direction, and applying Gauss's Law for the proper closed surface (a parallelepiped, with one side an imaginary plate to the left of the left plate and the other side an imaginary plate at distance x, between the two plates) we will conclude that the electric field is again independent of distance x and is equal to $\sigma/\epsilon_0$

In the case of the sperical capacitor the assumption is that the field will be in the radial direction which is the only valid assumption that can be made due to the spherical symmetry of the situation. From Gauss's Law we can conclude that the field will depend only on the charge of the inner sphere. The field due to the charge of the outer sphere is canceled out (the field inside would be zero if there was no inner charged sphere).

 

[Ras9]

I think I get it. So basically the field generated from a finite surface single plane is {\sigma \over 2\epsilon_0} only near the plane. If I move to far from it for symmetry reason I can not treat the field vectors as parallel and so the Gauss's law would give a different result (that I assume would be distance dependent). Is that right?

In the case of the sperical capacitor the assumption is that the field will be in the radial direction which is the only valid assumption that can be made due to the spherical symmetry of the situation. From Gauss's Law we can conclude that the field will depend only on the charge of the inner sphere. The field due to the charge of the outer sphere is canceled out (the field inside would be zero if there was no inner charged sphere).

Can you explain this more? I understand that the difference between spherical capacitor and plane capacitor is in the spherical symmetry of the first one, but if I think of Gauss's law would be natural to me to treat the external's sphere charge as a point charge situated at the center of the sphere. What am I missing?

 

[sunmaggot]

I think I get it. So basically the field generated from a finite surface single plane is {\sigma \over 2\epsilon_0} only near the plane. If I move to far from it for symmetry reason I can not treat the field vectors as parallel and so the Gauss's law would give a different result (that I assume would be distance dependent). Is that right?

Can you explain this more? I understand that the difference between spherical capacitor and plane capacitor is in the spherical symmetry of the first one, but if I think of Gauss's law would be natural to me to treat the external's sphere charge as a point charge situated at the center of the sphere. What am I missing?

In fact, for single plate, Gaussian surface is a cylindrical thing, ∫E ⋅ dA = Q/ε₀ → E = σ(πr²)/ε₀(2πr²) = σ/2ε₀
For capacitor, because both plates have same charges but different polarity, so the E field only travels inside the capacitor.
∫E⋅dA = Q/ε₀ → E=σ(πr²)/ε₀(πr²) = σ/ε₀
The difference is in fact the E flux. For single plate, E flux goes in both side. For capacitor, E flux only goes into one side for each plate.

 

[Delta2]

I think I get it. So basically the field generated from a finite surface single plane is {\sigma \over 2\epsilon_0} only near the plane. If I move to far from it for symmetry reason I can not treat the field vectors as parallel and so the Gauss's law would give a different result (that I assume would be distance dependent). Is that right?

yes right, where "near the plane" is defined such as the distance is much smaller than the two dimensions of the finite plane.

Can you explain this more? I understand that the difference between spherical capacitor and plane capacitor is in the spherical symmetry of the first one, but if I think of Gauss's law would be natural to me to treat the external's sphere charge as a point charge situated at the center of the sphere. What am I missing?

No it is not the same as having the external's sphere charge to a single point situated at the center of the sphere. What you missing is that if we remove the inner charged sphere, then the electric field inside the region of the external charged sphere will be zero (while if you had the charge as a point at center, the electric field in the same region would not be zero, however in the region outside the external charged sphere, the electric field would be the same as if the charge was at point at the center).

 

[Ras9]

No it is not the same as having the external's sphere charge to a single point situated at the center of the sphere. What you missing is that if we remove the inner charged sphere, then the electric field inside the region of the external charged sphere will be zero (while if you had the charge as a point at center, the electric field in the same region would not be zero, however in the region outside the external charged sphere, the electric field would be the same as if the charge was at point at the center).

Is this somehow connected that the external charge is created by induction from the internal charge?

 

[sunmaggot]

Is this somehow connected that the external charge is created by induction from the internal charge?

not accurate enough, its external charge is not created, it is just redistribution of charges. when external hollow sphere is neutral, you put a positive point charge (with +Q charge) in the centre of the hollow sphere. Then the inner surface of hollow sphere will have total of -Q charge. it is because the negative charges are attracted by the point charge in centre. However, remember one thing, the sphere is neutral initially, by conservation of charge, sphere's outer surface should have +Q charge. So, this is just redistribution, not induction at all.

 

[Delta2]

Is this somehow connected that the external charge is created by induction from the internal charge?

Why you mention the (electrostatic) induction, it sure plays a role on how the capacitor is charged or discharged , but your question wasnt that in first place. Your question was how the outer sphere doesn't contribute to the electric field of the spherical capacitor (while in the case of a parallel plate capacitor both plates contribute), am i right?

 

[Ras9]

not accurate enough, its external charge is not created, it is just redistribution of charges. when external hollow sphere is neutral, you put a positive point charge (with +Q charge) in the centre of the hollow sphere. Then the inner surface of hollow sphere will have total of -Q charge. it is because the negative charges are attracted by the point charge in centre. However, remember one thing, the sphere is neutral initially, by conservation of charge, sphere's outer surface should have +Q charge. So, this is just redistribution, not induction at all.

Yes I know, I have been a little superficial with my definition. Although I don't see any difference between induction and what you called redistribution.

Why you mention the (electrostatic) induction, it sure plays a role on how the capacitor is charged or discharged , but your question wasnt that in first place. Your question was how the outer sphere doesn't contribute to the electric field of the spherical capacitor (while in the case of a parallel plate capacitor both plates contribute), am i right?

Yes you are perfectly right, I thought that the two things might be connected.